Calculus

Fourier Series

Fourier Series Logo

Definition of Fourier Series and Typical Examples

  • Baron Jean Baptiste Joseph Fourier \(\left( 1768-1830 \right) \) introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related.

    Baron Jean Baptiste Joseph Fourier  (1768−1830)
    Fig.1 Baron Jean Baptiste Joseph Fourier (1768−1830)

    To consider this idea in more detail, we need to introduce some definitions and common terms.

    Basic Definitions

    A function \(f\left( x \right)\) is said to have period \(P\) if \(f\left( {x + P} \right) = f\left( x \right)\) for all \(x.\) Let the function \(f\left( x \right)\) has period \(2\pi.\) In this case, it is enough to consider behavior of the function on the interval \(\left[ { – \pi ,\pi } \right].\)

    1. Suppose that the function \(f\left( x \right)\) with period \(2\pi\) is absolutely integrable on \(\left[ { – \pi ,\pi } \right]\) so that the following so-called Dirichlet integral is finite:
      \[\int\limits_{ – \pi }^\pi {\left| {f\left( x \right)} \right|dx} \lt \infty ;\]
    2. Suppose also that the function \(f\left( x \right)\) is a single valued, piecewise continuous (must have a finite number of jump discontinuities), and piecewise monotonic (must have a finite number of maxima and minima).

    If the conditions \(1\) and \(2\) are satisfied, the Fourier series for the function \(f\left( x \right)\) exists and converges to the given function (see also the Convergence of Fourier Series page about convergence conditions.)

    At a discontinuity \({x_0}\), the Fourier Series converges to

    \[\lim\limits_{\varepsilon \to 0} \frac{1}{2}\left[ {f\left( {{x_0} – \varepsilon } \right) – f\left( {{x_0} + \varepsilon } \right)} \right].\]

    The Fourier series of the function \(f\left( x \right)\) is given by

    \[{f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}} ,}\]

    where the Fourier coefficients \({{a_0}},\) \({{a_n}},\) and \({{b_n}}\) are defined by the integrals

    \[
    {{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} ,\;\;\;}\kern-0.3pt
    {{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nx dx} ,\;\;\;}\kern-0.3pt
    {{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nx dx} .}
    \]

    Sometimes alternative forms of the Fourier series are used. Replacing \({{a_n}}\) and \({{b_n}}\) by the new variables \({{d_n}}\) and \({{\varphi_n}}\) or \({{d_n}}\) and \({{\theta_n}},\) where

    \[
    {{d_n} = \sqrt {a_n^2 + b_n^2} ,\;\;\;}\kern-0.3pt
    {\tan {\varphi _n} = \frac{{{a_n}}}{{{b_n}}},\;\;\;}\kern-0.3pt
    {\tan {\theta _n} = \frac{{{b_n}}}{{{a_n}}},}
    \]

    we can write:

    \[ {f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{d_n}\sin \left( {nx + {\varphi _n}} \right)} \;\;}\kern-0.3pt{\text{or}\;\;} {f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{d_n}\cos\left( {nx + {\theta _n}} \right)} .} \]

    Fourier Series of Even and Odd Functions

    The Fourier series expansion of an even function \(f\left( x \right)\) with the period of \(2\pi\) does not involve the terms with sines and has the form:

    \[{f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{a_n}\cos nx} ,}\]

    where the Fourier coefficients are given by the formulas

    \[
    {{a_0} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)dx} ,\;\;\;}\kern-0.3pt
    {{a_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\cos nxdx} .}
    \]

    Accordingly, the Fourier series expansion of an odd \(2\pi\)-periodic function \(f\left( x \right)\) consists of sine terms only and has the form:

    \[f\left( x \right) = \sum\limits_{n = 1}^\infty {{b_n}\sin nx} ,\]

    where the coefficients \({{b_n}}\) are

    \[{b_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\sin nxdx} .\]

    Below we consider expansions of \(2\pi\)-periodic functions into their Fourier series, assuming that these expansions exist and are convergent.


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Let the function \(f\left( x \right)\) be \(2\pi\)-periodic and suppose that it is presented by the Fourier series:
    \[{f\left( x \right) = \frac{{{a_0}}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}}}\]
    Calculate the coefficients \({{a_0}},\) \({{a_n}},\) and \({{b_n}}.\)

    Example 2

    Find the Fourier series for the square \(2\pi\)-periodic wave defined on the interval \(\left[ { – \pi ,\pi } \right]:\)
    \[ {f\left( x \right) \text{ = }}\kern0pt {\begin{cases} 0, & \text{if} & – \pi \le x \le 0 \\ 1, & \text{if} & 0 < x \le \pi \end{cases}.} \]

    Example 3

    Find the Fourier series for the sawtooth wave defined on the interval \(\left[ { – \pi ,\pi } \right]\) and having period \(2\pi.\)

    Example 4

    Let \(f\left( x \right)\) be a \(2\pi\)-periodic function such that \(f\left( x \right) = {x^2}\) for \(x \in \left[ { – \pi ,\pi } \right].\) Find the Fourier series for the parabolic wave.

    Example 5

    Find the Fourier series for the triangle wave
    \[ {f\left( x \right) \text{ = }}\kern0pt {\begin{cases} \frac{\pi }{2} + x, & \text{if} & – \pi \le x \le 0 \\ \frac{\pi }{2} – x, & \text{if} & 0 \lt x \le \pi \end{cases},} \]
    defined on the interval \(\left[ { – \pi ,\pi } \right].\)

    Example 6

    Find the Fourier series for the function
    \[ {f\left( x \right) \text{ = }}\kern0pt {\begin{cases} -1, & \text{if} & – \pi \le x \le – \frac{\pi }{2} \\ 0, & \text{if} & – \frac{\pi }{2} \lt x \le \frac{\pi }{2} \\ 1, & \text{if} & \frac{\pi }{2} \lt x \le \pi \end{cases},} \]
    defined on the interval \(\left[ { – \pi ,\pi } \right].\)

    Example 1.

    Let the function \(f\left( x \right)\) be \(2\pi\)-periodic and suppose that it is presented by the Fourier series:
    \[{f\left( x \right) = \frac{{{a_0}}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}}}\]
    Calculate the coefficients \({{a_0}},\) \({{a_n}},\) and \({{b_n}}.\)

    Solution.

    To define \({{a_0}},\) we integrate the Fourier series on the interval \(\left[ { – \pi ,\pi } \right]:\)

    \[
    {\int\limits_{ – \pi }^\pi {f\left( x \right)dx} }
    = {\pi {a_0} }+{ \sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ – \pi }^\pi {\cos nxdx} }\right.}+{\left.{ {b_n}\int\limits_{ – \pi }^\pi {\sin nxdx} } \right]}}
    \]

    For all \(n \gt 0\),

    \[ {{\int\limits_{ – \pi }^\pi {\cos nxdx} }={ \left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_{ – \pi }^\pi }={ 0\;\;}{\text{and}\;\;\;}}\kern-0.3pt {{\int\limits_{ – \pi }^\pi {\sin nxdx} }={ \left. {\left( { – \frac{{\cos nx}}{n}} \right)} \right|_{ – \pi }^\pi }={ 0.}} \]

    Therefore, all the terms on the right of the summation sign are zero, so we obtain

    \[
    {\int\limits_{ – \pi }^\pi {f\left( x \right)dx} = \pi {a_0}\;\;\text{or}\;\;\;}\kern-0.3pt
    {{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} .}
    \]

    In order to find the coefficients \({{a_n}},\) we multiply both sides of the Fourier series by \(\cos mx\) and integrate term by term:

    \[ {\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} } = {\frac{{{a_0}}}{2}\int\limits_{ – \pi }^\pi {\cos mxdx} } + {\sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} }\right.}}+{{\left.{ {b_n}\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} } \right]} .} \]

    The first term on the right side is zero. Then, using the well-known trigonometric identities, we have

    \[
    {\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} }
    = {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\sin{\left( {n + m} \right)x} }\right.}+{\left.{ {\sin \left( {n – m} \right)x}} \right]dx} }={ 0,}
    \]

    \[
    {\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} }
    = {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\cos {\left( {n + m} \right)x} }\right.}+{\left.{ {\cos \left( {n – m} \right)x}} \right]dx} }={ 0,}
    \]

    if \(m \ne n.\)

    In case when \(m = n\), we can write:

    \[\require{cancel}
    {\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} }
    = {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\sin 2mx + \sin 0} \right]dx} ,\;\;}\Rightarrow
    {\int\limits_{ – \pi }^\pi {{\sin^2}mxdx} }={ \frac{1}{2}\left[ {\left. {\left( { – \frac{{\cos 2mx}}{{2m}}} \right)} \right|_{ – \pi }^\pi } \right] }
    = {\frac{1}{{4m}}\left[ { – \cancel{\cos \left( {2m\pi } \right)} }\right.}+{\left.{ \cancel{\cos \left( {2m\left( { – \pi } \right)} \right)}} \right] }={ 0;}
    \]

    \[
    {\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} }
    = {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\cos 2mx + \cos 0} \right]dx} ,\;\;}\Rightarrow
    {\int\limits_{ – \pi }^\pi {{\cos^2}mxdx} }
    = {\frac{1}{2}\left[ {\left. {\left( {\frac{{\sin 2mx}}{{2m}}} \right)} \right|_{ – \pi }^\pi + 2\pi } \right] }
    = {\frac{1}{{4m}}\left[ {\sin \left( {2m\pi } \right) }\right.}-{\left.{ \sin \left( {2m\left( { – \pi } \right)} \right)} \right] + \pi }={ \pi .}
    \]

    Thus,

    \[
    {\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} = {a_m}\pi ,\;\;}\Rightarrow
    {{a_m} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} ,\;\;}\kern-0.3pt
    {m = 1,2,3, \ldots }
    \]

    Similarly, multiplying the Fourier series by \(\sin mx\) and integrating term by term, we obtain the expression for \({{b_m}}:\)

    \[
    {{b_m} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin mxdx} ,\;\;\;}\kern-0.3pt
    {m = 1,2,3, \ldots }
    \]

    Rewriting the formulas for \({{a_n}},\) \({{b_n}},\) we can write the final expressions for the Fourier coefficients:

    \[
    {{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nxdx} ,\;\;\;}\kern-0.3pt
    {{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nxdx} .}
    \]

    Example 2.

    Find the Fourier series for the square \(2\pi\)-periodic wave defined on the interval \(\left[ { – \pi ,\pi } \right]:\)
    \[ {f\left( x \right) \text{ = }}\kern0pt {\begin{cases} 0, & \text{if} & – \pi \le x \le 0 \\ 1, & \text{if} & 0 < x \le \pi \end{cases}.} \]

    Solution.

    First we calculate the constant \({{a_0}}:\)

    \[
    {{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} }
    = {\frac{1}{\pi }\int\limits_0^\pi {1dx} }
    = {\frac{1}{\pi } \cdot \pi }={ 1.}
    \]

    Find now the Fourier coefficients for \(n \ne 0:\)

    \[
    {{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nxdx} }
    = {\frac{1}{\pi }\int\limits_0^\pi {1 \cdot \cos nxdx} }
    = {\frac{1}{\pi }\left[ {\left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_0^\pi } \right] }
    = {\frac{1}{{\pi n}} \cdot 0 }={ 0,}
    \]

    \[
    {{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nxdx} }
    = {\frac{1}{\pi }\int\limits_0^\pi {1 \cdot \sin nxdx} }
    = {\frac{1}{\pi }\left[ {\left. {\left( { – \frac{{\cos nx}}{n}} \right)} \right|_0^\pi } \right] }
    = { – \frac{1}{{\pi n}} \cdot \left( {\cos n\pi – \cos 0} \right) }
    = {\frac{{1 – \cos n\pi }}{{\pi n}}.}
    \]

    As \(\cos n\pi = {\left( { – 1} \right)^n},\) we can write:

    \[{b_n} = \frac{{1 – {{\left( { – 1} \right)}^n}}}{{\pi n}}.\]

    Thus, the Fourier series for the square wave is

    \[{f\left( x \right) = \frac{1}{2} }+{ \sum\limits_{n = 1}^\infty {\frac{{1 – {{\left( { – 1} \right)}^n}}}{{\pi n}}\sin nx} .}\]

    We can easily find the first few terms of the series. By setting, for example, \(n = 5,\) we get

    \[ {f\left( x \right) = \frac{1}{2} }+{ \frac{{1 – \left( { – 1} \right)}}{\pi }\sin x } + {\frac{{1 – {{\left( { – 1} \right)}^2}}}{{2\pi }}\sin 2x } + {\frac{{1 – {{\left( { – 1} \right)}^3}}}{{3\pi }}\sin 3x } + {\frac{{1 – {{\left( { – 1} \right)}^4}}}{{4\pi }}\sin 4x } + {\frac{{1 – {{\left( { – 1} \right)}^5}}}{{5\pi }}\sin 5x + \ldots } = {\frac{1}{2} + \frac{2}{\pi }\sin x } + {\frac{2}{{3\pi }}\sin 3x } + {\frac{2}{{5\pi }}\sin 5x + \ldots } \]

    The graph of the function and the Fourier series expansion for \(n = 10\) is shown below in Figure \(2.\)

    Fourier series for the square  2pi-periodic wave
    Figure 2, n = 10
    Page 1
    Problems 1-2
    Page 2
    Problems 3-6